114年屏東高中教甄-數學詳解

 國立屏東高中 114 學年第 1 次正式教師甄試

一、填充題:每題 5 分，共 50 分

解答:$$f(x)=x^5+x^4-x^2+1=(x-\alpha_1)(x-\alpha_2) (x-\alpha_3) (x-\alpha_4)(x-\alpha_5)=0\\ \Rightarrow f(1) =(1-\alpha_1) (1-\alpha_2)(1-\alpha_3)(1-\alpha_4) (1-\alpha_5) =0 \\ P(x)=x^4-1=(x^2+1)(x^2-1)=(x+i)(x-1)(x+1)(x-1) =(x-1)Q(x) \\ P(\alpha_1)\times P(\alpha_2) \times \cdots \times P(\alpha_5) =(\alpha_1-1)Q(\alpha_1) \times (\alpha_2-1)Q(\alpha_2)\times \cdots \times (\alpha_5-1)Q(\alpha_5) \\=-(1-\alpha_1)(1-\alpha_2)(1-\alpha_3)(1-\alpha_4) (1-\alpha_5) Q(\alpha_1) Q(\alpha_2 )Q(\alpha_3 )Q(\alpha_4 )Q(\alpha_5) \\=-f(1) Q(\alpha_1) Q(\alpha_2)Q(\alpha_3 )Q(\alpha_4 )Q(\alpha_5) =\bbox[red, 2pt]0$$

解答:$$取\cases{u=13x-9y+5\\ v=9x+13y-4} \Rightarrow u^2+v^2\le 1000 \Rightarrow 圓面積=1000\pi\\ {\partial(u,v) \over \partial (x,y)} = \begin{Vmatrix} u_x & u_y\\ v_x& v_y\end{Vmatrix} =\begin{Vmatrix} 13 & -9\\ 9& 13\end{Vmatrix}=13^2+9^2=250 \Rightarrow 欲求之面積={1000\pi\over 250} =\bbox[red, 2pt]{4\pi}$$

解答:$$\cases{y=f(x)=2e^x\\ y=g(x)=\ln(x/2)} \Rightarrow f=g^{-1} \Rightarrow 兩圖形對稱於直線L:y=x\\ \Rightarrow \overline{PQ}最小值=2\times d(L,y=g(x))的最小值\\ d(L,y=g(x))=h(x) ={|\ln (x/2)-x|\over \sqrt 2} \Rightarrow h'(x)=0 \Rightarrow {1\over x}-1=0 \Rightarrow x=1 \\\Rightarrow h(1)={1\over \sqrt 2}(1+\ln 2) \Rightarrow \overline{PQ}最小值=2h(1)= \bbox[red, 2pt]{\sqrt 2(1+\ln 2)}$$

解答:$${k\over C^{k+2}_3} ={k\over {(k+2)(k+1)k\over 3!}}= {6\over (k+2)(k+1)}=6 \left({1\over k+1}-{1\over k+2} \right) \\ \Rightarrow \sum_{n=1}^\infty {n\over C^{n+2}_3} =6\sum_{n=1}^\infty \left({1\over k+1}-{1\over k+2} \right) =6\cdot {1\over 2}= \bbox[red, 2pt]3$$

解答:$$假設\cases{x^2-x+a=0 的兩根為\alpha, \beta\\ x^2-x+b=0的兩根為\gamma, \delta} \Rightarrow \cases{\alpha+ \beta=1\\ \gamma+\delta=1} \Rightarrow 等差數列\langle a_n\rangle 滿足\cases{a_1+a_2+a_3+a_4= 2\\ a_1=1/4} \\ \Rightarrow 2(2a_1+3d) =2({1\over 2}+3d)= 2\Rightarrow d= \bbox[red, 2pt]{1\over 6}$$

解答:$$假設欲求之直線L: y=m(x-4)+2及交點\cases{A(x_1,y_1) \\B(x_2,y_2)} \Rightarrow (4,2)=(A+B)/2 \Rightarrow \cases{x_1+x_2=8\\ y_1+y_2=4} \\ 將y=m(x-4)+2代入雙曲線\Rightarrow 4x^2-(m(x-4)+2)^2=9 \\\Rightarrow (4-m^2)x^2+2m(4m-2)x-(2-4m)^2-9=0 \Rightarrow 兩根之和=x_1+x_2={2m(4m-2) \over m^2-4} =8 \\ \Rightarrow {8m^2-4m\over m^2-4}=8 \Rightarrow 4m=32 \Rightarrow m=8 \Rightarrow L:y=8(x-4)+2 \Rightarrow \bbox[red, 2pt]{8x-y=30}$$

解答:$$假設\cases{\overline{BC}=a \\ \overline{AC}=b} \Rightarrow \cos \angle ACB=\cos 60^\circ ={1\over 2}={a^2+b^2-49\over 2ab} \Rightarrow a^2+b^2-ab=49 \\ \Rightarrow \cases{a=1 \Rightarrow b^2-b-48=0 \Rightarrow b \not \in \mathbb N\\ a=2 \Rightarrow b^2-2b-45=0 \Rightarrow b \not \in \mathbb N\\ a=3 \Rightarrow b^2-3b-40=0 \Rightarrow b=8\\ a=4 \Rightarrow b^2-4b-33=0 \Rightarrow b \not \in \mathbb N\\ a=5 \Rightarrow b^2-5b-24=0 \Rightarrow b=8\\ a=6 \Rightarrow b^2-6b-13=0\Rightarrow b \not \in \mathbb N\\a=7 \Rightarrow b^2-7b=0 \Rightarrow b=7\\ a=8 \Rightarrow b^2-8b+15 \Rightarrow b=3,5\\ a=9 \Rightarrow b^2-9b+32=0\Rightarrow 判別式\lt 0\\ a=10 \Rightarrow 判別式\lt 0\\ \cdots \cdots } \Rightarrow (a,b)=(3,8),(5,8),(7,7),(8,3), (8,5) \\ \Rightarrow 共\bbox[red, 2pt]{5}個三角形$$

解答:$$假設\cases{Z_1=a_1+b_1i\\ Z_2=a_2+b_2i\\ Z_3=a_3+b_3i} ,a_i,b_i\in \mathbb R \Rightarrow \cases{|Z_1|=\sqrt 2\\ |Z_2|=\sqrt 5\\ |Z_3|=3\\ O為\triangle PQR重心} \Rightarrow \cases{ a_1^2+ b_1^2=2 \cdots(1)\\ a_2^2+b_2^2 =5 \cdots(2)\\ a_3^2+b_3^2=9 \cdots(3)\\ a_1+a_2+a_3=0\\ b_1+b_2+b_3 =0} \\ (1)+(2) =a_1^2+a_2^2 +b_1^2+b_2^2 =7 \Rightarrow (a_1+a_2)^2-2a_1a_2 +(b_1+b_2)^2-2b_1b_2=7 \\ \Rightarrow (-a_3)^2+ (-b_3)^2-2(a_1a_2+b_1b_2) =9-2(a_1a_2+b_1b_2)=7 \Rightarrow a_1a_2 +b_1b_2=1\\ \Rightarrow Re(\overline{Z_1} \cdot Z_2)= Re((a_1-b_1i)(a_2+b_2i)) =a_1a_2+b_1b_2 =\bbox[red, 2pt]  1$$

解答:$$第8次才停止的情形:\square\square\square\square\square 正 正正,有2^5=32種情形，但需扣除以下情形\\ \cases{\square \square\square\square正:有2^4=16種\\ \square正正正反:有2種\\ 正正正反反:有1種} \Rightarrow 合計19種\Rightarrow 第8次才停止共有32-19=13種\\ 其中第一次是正面的情形: 正正反反反,正正反正反, 正反\square \square 反,共1+1+4=6 \\ 因此條件機率= \bbox[red, 2pt]{6\over 13}$$

解答:$$\cases{A(0,0,0) \\B(a,0,a)\\ C(0,a,a)\\ D(a,a,0)} \Rightarrow 邊長\overline{AB}=\sqrt 2 a=9 \Rightarrow a=9/\sqrt 2 \\ \Rightarrow \cases{平面E_1=\triangle BCD: x+y+z=2a\\ 平面E_2=\triangle ACD: x-y+z=0} \Rightarrow \cases{A'(4a/3,4a/3, 4a/3) \\B'(-a/3,4a/3,-a/3)} \\ \Rightarrow \overline{A'B'} =\sqrt 2\cdot (5a/3) =\sqrt 2\cdot {5\over 3}\cdot {9\over \sqrt 2} =\bbox[red, 2pt]{15}$$

二、計算題:每題 10 分，共 50 分

解答:$$\textbf{(1) }y=x^3 \Rightarrow y'=3x^2 \Rightarrow 過切點P(a,a^3)之切線L: y=3a^2(x-a)+a^3 \\ 當x=0 \Rightarrow y=-3a^3+a^3=-2a^3 \Rightarrow Q(0,-2a^3) \\將y=3a^2(x-a)+a^3代入\Gamma \Rightarrow 3a^2(x-a)+a^3=x^3 \Rightarrow x^3-3a^2x+2a^3=0 \\ \Rightarrow (x-a)(x^2+ax-2a^2) =(x-a)^2(x+2a)=0 \Rightarrow x=-2a \Rightarrow S(-2a,-8a^3) \\ \Rightarrow {\overline{PQ} : \overline{QS}} =\sqrt{a^2+(a^3+2a^3)^2}: \sqrt{4a^2+(8a^3-2a^3)^2} =\sqrt{a^2+9a^6} :\sqrt{4a^2+36a^6} =\bbox[red, 2pt]{1:2} \\\textbf{(2) }所圍面積 \int_{-2a}^a x^3-(3a^2(x-a)+a^3) \,dx = \int_{-2a}^a x^3-3a^2x +2a^3 \,dx \\=  \left. \left[{1\over 4}x^4-{3\over 2}a^2x^2 +2a^3x \right] \right|_{-2a}^a = \bbox[red, 2pt]{{27\over 4}a^4}$$

解答:

$$過A(2,6,-3)且方向向量為(1,2,2)的直線L:{x-2\over 1}={y-6\over 2} ={z+3\over 2}\\ A在平面E的投影點A'(t+2,2t+6,2t-3) \Rightarrow t+2+2(2t+6)+2(2t-3)+1=0 \Rightarrow t=-1\\ \Rightarrow A'(1,4,-5) \Rightarrow \overline{A'Q} =6 \gt 2(圓半徑r)\Rightarrow A'在圓外 \\假設\overleftrightarrow{A'Q}與圓交於B(離A'較近),C(離A'較遠)兩點 ,則\overline{AP}的最大值=\overline{AC}, 最小值=\overline{AB} \\ \Rightarrow \cases{\overline{AA'}= \sqrt{1+4+4}=3\\ \overline{A'B} =\overline{A'Q}-r=4 \\ \overline{A'C}= \overline{A'Q}+ r=8}\Rightarrow \cases{\overline{AC}^2= \overline{A'C}^2 +\overline{AA'}^2 =64+9=73\\ \overline{AB}^2 =\overline{A'B}^2 +\overline{AA'}^2 =16+9=25} \Rightarrow \bbox[red, 2pt]{ \cases{最大值:\sqrt{73} \\ 最小值:5}}$$

解答:$$\cases{L_1方向向量\vec u=(1,2,-2) \\ L_2方向向量\vec v =(3,1,-2)} \Rightarrow \vec n=\vec u\times \vec v=(-2,-4,-5)\\ \Rightarrow 包含L_2的平面E:-2x-4(y-2)-5(z+1)=0 \Rightarrow 2x+4y+5z=3\\ P(3,0,-2) \in L_1 \Rightarrow d(P,E)={|6+0-10-3|\over \sqrt{4+16+25}} ={7\over 3\sqrt 5}=正\triangle PQR的高\\ \Rightarrow \triangle PQR面積= \left( {7\over 3\sqrt 5}\right)^2\cdot {1\over \sqrt 3} ={49\over 45}\cdot {1\over \sqrt 3} = \bbox[red, 2pt]{49\sqrt 3\over 135}$$

解答:$$f(A,B,C)= \cos^2(A-B)+ \cos^2(B-C)+ \cos^2(C-A) \\ \Rightarrow f_A=0 \Rightarrow -2\cos(A-B)\sin(A-B)+2 \cos(C-A)\sin(C-A) =0 \Rightarrow\sin(2A-2B)= \sin(2C-2A) \\ \Rightarrow A=(B+C)/2 \Rightarrow f((B+C)/2,B,C) =\cos^2(C/2-B/2)+ \cos^2(B-C) +\cos^2(C/2-B/2) \\=2 \cos^2(B/2-C/2)+ \cos^2(B-C) =2\cos(B-C)+1+\cos^2(B-C) =(\cos(B-C)+{{1\over 2}})^2+{3\over 4} \\ \Rightarrow f(A,B,C)的最小值=\bbox[red, 2pt]{3\over 4}$$

解答:$$p ={1\over 4}p+{1\over 4}p^2+ {1\over 4}p^3+{1\over 4} \Rightarrow p^3+p^2-3p+1=0 \Rightarrow (p-1)(p^2+2p-1)=0 \\ \Rightarrow p^2+2p-1=0 \Rightarrow p={-2+2\sqrt 2\over 2} = \bbox[red, 2pt]{\sqrt 2-1}$$

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